Question

The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = A cos (ωt + φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Answer 1

Initially, at t = 0:

Displacement, x = 1 cm

Initial velocity, v = ω cm/sec.

Angular frequency, ω = π rad/s^–1

It is given that:

`x(t) = A cos(omegat + phi)`

`1 = A cos(omega xx 0 xx phi)`

`A cos phi = 1`   ....(i)

`"Velocity", v = (dx)/(dt)`

`omega = - Aomegasin(omegat +  phi)`

`1 = -A sin (omega xx 0 + phi)` = `- A Sin phi`

`Asin phi = -1`  ...(ii)

Squaring and adding equations (i) and (ii), we get:

`A^2(sin^2 phi + cos^2 phi) = 1+ 1`

`A^2= 2`

`:. A = sqrt2 cm`

Dividing equation (ii) by equation (i), we get:

`tan phi = -1`

`: phi = (3pi)/4, (7pi)/4,....`

SHM is given as:

`x = B sin(omegat + alpha)`

Putting the given values in this equation, we get:

`1 = B sin[omegaxx0 + alpha]`

`B sin alpha = 1`  ....(iii)

Velocity, `v = omegaBcos(omegat + alpha)`

Substituting the given values, we get:

`pi = piBsin alpha`

`Bsin alpha = 1`       .....(iv)

Squaring and adding equations (iii) and (iv), we get:

`B^2[sin^2alpha + cos^2 alpha] = 1+ 1`

`B^2 =  2`

`:. B = sqrt2 cm`

Dividing equation (iii) by equation (iv), we get:

`(Bsin alpha)/(B cos alpha) = 1/1`

`tan alpha = 1 = tan pi/4`

`:. alpha = pi/4 , (5pi)/4, ......`

 

Answer 2

The given displacement function is 

`x(t) =  A cos (omegat + phi)`   ....(i)

At t = 0, x(0) =  1 cm,Also `omega = pis^(-1)` 

`:. 1 = A cos (pi xx 0 +phi)`

`=> A cos phi = 1`

Also, differentiating equation (i)  w.r.t 't'

`v= d/(dt) x(t ) = -A omega sin (omegat + phi)` 

Now at t = 0, `v = omega`

:. From equation  (iii) `omega =  -A omega sin (pi xx0+ phi)` or `A sinphi =  -1`

Squaring and adding eqations (ii) and (iv)

`A^2 cos^2phi + A^2 sin^2  phi = 1^2 + 1^2` or  `A = sqrt2 cm`

Dividing equation (ii) and (iv)

`(A sin phi)/(A cos phi) = (-1)/1`

`:. tan phi = -1 => phi = (3pi)/4`

If instead we use the sine function, i.e

`x = Bsin (omegat + alpha)` ,then `v = d/(dt) B omega cos (omegat + alpha)`

:. At t = 0 using x= 1  and  `v =  omega`, we get `1  = B sin(omega xx 0 + alpha)`

or `B sin alpha = 1`  ....(v)

and `omega =  Bomega cos (omega xx 0 + alpha)`  or `B cos alpha =  1` ...(vi)

Dividing v and vi

`tan alpha = 1 or alpha = pi/4 or (5pi)/4`

Squaring v and vi, we get

`B^2 sin^2 alpha + B^2 cos^2 alpha  = 1^2 + 1^2`

`=> B =  sqrt2 cm`